Q109 / 120 · NDA 2024 (Phase 2) · Mathematics

A natural number $x$ is chosen at random from the first 100 natural numbers. What is the probability that $x^2+x>50$ ?

  • A
    $\frac{93}{100}$
  • Correct answer
    B
    $\frac{47}{50}$
  • C
    $\frac{24}{25}$
  • D
    $\frac{23}{25}$

Solution

First 100 natural numbers $\mathrm{n}(\mathrm{~s})=100$ Number from 7 to 100 satisfy $\begin{aligned} & x^2+x>50 \\ & \therefore \quad n(E)=100-6=94 \\ & P(E)=\frac{n(E)}{n(S)}=\frac{94}{100}=\frac{47}{50} \end{aligned}$

Hence, the correct answer is option: 'B'.

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